Math 20A - Calculus by Jon Rogawski Chapter 4 - Applications of the derivative Prepared by Jason Gaddis 1 Linear approximation and applications Remark 1.1. We can use tangent lines to approximate a curve at a specific point. If we allow for a small amount of error, we can approximate the curve on some interval The error grows as we make the interval larger. This is called linear approximation. Definition 1.2. If f is differentiable at x = a and ∆x is small, then ∆f ≈ f 0(a)∆x where ∆x = f(a+δx)−f(a) is the linear approximation of ∆f (or tangent line approximation). p Example 1.3. Use Linear approximation to estimate 100:2 − 10. p 0 1 −1=2 Let f(x) = x at x = 100 and ∆x = :2. Then f (x) = 2x , so ∆f is approximated by 1 1 f 0(100)∆x = (:02) = :001: 2 10 p A calculator gives 100:2 = 10:009995005, so the error in the difference is j009995005 − :001j = 0:008995005: Remark 1.4 (Differential notation). If y = f(x), where f is a differentiable function, then the differential dx is an independent variable. The differential dy is then defined in terms of dx by the equation dy = f 0(a)dx: Now, ∆y ≈ dy. Definition 1.5. If f is differentiable and x is close to a then the lineariza- tion of f at a is the linear function whose graph is the tangent line, that is, L(x) = f(a) + f 0(a)(x − a): 1 π Example 1.6. Find the linearization of the function f(x) = sin x at a = 6 π and use it to approximate sin( 5 ). p 1 0 3 We have sin(π=6) = 2 and f (π=6) = cos(π=6) = 2 . Then the linearization of f at a is p 1 3 π L(x) = + (x − ): 2 2 6 Then L(π=5) = 0:591. The actual value 0:588. Remark 1.7. Error is relatively meaningless by iteslf. To standardize error, we define the percentage error to be error × 100%: actual error Example 1.8. Compute the percentage error in the previous example. The actual value was 0:588, so the error is j:591 − :588j = :003. Hence, the percentage error is :003 × 100 = :510: :588 1 Thus, the error is approximately 2 of a percent. 2 2 Extreme Values Definition 2.1. Let f(x) be a function on an interval I and let a 2 I. We say that f(a) is the absolute maximum of f on I if f(x) ≤ f(a) for all x 2 I. We say that f(a) is the absolute minimum of f on I if f(x) ≥ f(a) for all x 2 I. Remark 2.2. The word \absolute" is sometimes replaced by \global" Remark 2.3. A function may not necessarily have an absolute maximum or minimum on some interval. Consider, for example, the function f(x) = x2 on any open interval. However, there is one case where we can guarantee the existence of absolute extrema. Theorem 2.4 (The Extreme Value Theorem). If f is continuous on a closed interval [a; b], then f attains an absolute maximum and absolute minimum on [a; b]. Remark 2.5. If either of the hypotheses above are removed, the theorem is no longer holds. Definition 2.6 (Local Extrema). A function f has a local maximum at x = c if f(c) ≥ f(x) for all x in some open interval containing c. Similarly, f has a local minimum at c if f(c) ≤ f(x) for all x in some open interval containing c. Remark 2.7. We can see relatively easily from a graph that local extrema occur at points where the graph of the function has a horizontal tangent line or where the derivative does not exist (vertical tangent line). Definition 2.8. A number c in the domain of f is a critical number if f 0(c) = 0 or f 0(c) does not exist. Example 2.9. Find the critical numbers of f(x) = x3 − x2 − x. We have f 0(x) = 3x2 − 2x − 1 = (3x + 1)(x − 1). Thus, f 0(x) = 0 when x = −1=3 or x = 1. These are the critical points of f. Theorem 2.10 (Fermat's Theorem). If f has a local maximum or minimum at c, then c is a critical point. 3 Remark 2.11. This is only one of many theorems attributed to Pierre de Fermat (1601-1665), who was a lawyer and amateur mathematician. Proof. Suppose f(c) is a local minimum. If f 0(c) does not exist, we are done, so assume it does exist. We claim f 0(c) = 0. Because f(c) is a local minimum, f(c + h) ≥ f(c) for small values of h 6= 0. That is, f(c + h) − f(c) ≥ 0. Then f(c + h) − f(c) f 0(c) = lim ≥ 0 h!0+ h f(c + h) − f(c) f 0(c) = lim ≤ 0: h!0− h Hence, f 0(c) ≥ 0 and f 0(c) ≤ 0, but this only holds if f 0(c) = 0. Remark 2.12. Consider f(x) = x3. This has a critical point at x = 0 but it is not a local max or min. Thus, while all local max and min occur at critical points, not all critical points yield local extrema. Example 2.13. In the previous example, determine whether the critical points are local max, local min, or neither. We have f(x) = x3 − x2 − x, f 0(x) = 3x2 − 2x − 1 = (3x + 1)(x − 1), and critical points c = −1=3; x = 1. For x < −1=3, f 0(x) > 0, so the function is increasing. For −1=3 < x < 1, f 0(x) < 0, so the function is decreasing. Since the function switched from increasing to decreasing at x = −1=3, it is a local max. Similarly, for x > 1, f 0(x) > 0, so the function is increasing. Since the function switched from decreasing to increasing at x = 1, it is a local min. Theorem 2.14 (Extreme values on a closed interval). Assume that f(x) is continuous on [a; b] and let f(c) be the minimum or maximum value on [a; b]. Then c is either a critical point or one of the endpoints a or b. Example 2.15. Find the abs. max and abs. min values of f(x) = x3 −6x2 + 9x + 2 on the closed interval [−1; 4]. Theorem 2.16 (Rolle's Theorem). Assume that f(x) is continuous on [a; b] and differentiable on (a; b). If f(a) = f(b), then there exists a number c between a and b such that f 0(c) = 0. 4 Example 2.17. Verify that the function f(x) = x3 −3x2 +2x+5 satisfies the hypotheses of Rolle's Theorem on the interval [0; 2]. Then find all numbers c that satisfy the conclusion of the theorem. 5 3 The Mean Value Theorem and Monotonicity Theorem 3.1 (Rolle's Theorem). Assume that f(x) is continuous on [a; b] and differentiable on (a; b). If f(a) = f(b), then there exists a number c between a and b such that f 0(c) = 0. Example 3.2. Verify that the function f(x) = x3 − 3x2 + 2x + 5 satisfies the hypotheses of Rolle's Theorem on the interval [0; 2]. Then find all numbers c that satisfy the conclusion of the theorem. Example 3.3. Show that the equation 2x − 1 − sin x = 0 has exactly one real root. Theorem 3.4 (The Mean Value Theorem). Let f be a function that satisfies the following hypotheses: 1. f is continuous on the closed interval [a; b] 2. f is differentiable on the open interval (a; b) Then there is a number c in (a; b) such that f(b) − f(a) f 0(c) = : b − a Equivalently, f(b) − f(a) = f 0(c)(b − a): Example 3.5. Verify that the function f(x) = x3 + x − 1 satisfies the hy- potheses of the Mean Value Theorem on the interval [0; 2]. Then find all numbers c that satisfy the conclusion of the theorem. Theorem 3.6. If f 0(x) = 0 for all x in the interval (a; b), then f is constant on (a; b). Corollary 3.7. If f 0(x) = g0(x) for all x in an interval (a; b), hen f − g is constant on (a; b); that is, f(x) = g(x) + c where c is a constant. Remark 3.8. We will see how the derivative gives us information about the behavior of a function. Our first result is a consequence of the MVT. Theorem 3.9 (The sign of the derivative). Let f be differentiable on an open interval (a; b). 6 (a) If f 0(x) > 0 on an interval, then f is increasing on that interval. (b) If f 0(x) < 0 on an interval, then f is decreasing on that interval. 0 Proof. Suppose f (x) > 0 for all x 2 (a; b). Choose points x1 < x2 in (a; b). We claim f(x2) > f(x1). By the MVT, there exists a c 2 (x1; x2) such that 0 f(x2) − f(x1) = f (c)(x2 − x1) > 0: 0 Therefore, f(x2) > f(x1). The case f (x) < 0 is similar. Example 3.10. Given f(x) = x4 − 4x − 1, find the intervals on which f is increasing or decreasing.